Question

my code is as follows:-

#include<stdio.h>

main() {
    union a {
        short int x;
        char y[2];
    };
    union a e;
    e.y[0] = 3;
    e.y[1] = 2;
    printf("%d\n%d\n%d\n", e.y[0], e.y[1], e.x);

    return 0;
}

It gives output as

3
2
515

i didn’t understand that how this 515 comes?

Answer 1

Since I wanted an excuse for ASCII art…

Declaring the union lets you choose how to interpret its data. In your case, either as an unsigned short or a char [2]. Both of these are 2 bytes long, so your union will refer to a 2-byte section of memory, thusly:

union a e;

    ---------------
e-> | ???? | ???? |
    ---------------

Now you decide to interpret your union as a character array:

e.y[0] = 3;

    ---------------
e-> | 0x03 | ???? |
    ---------------

e.y[1] = 2;

    ---------------
e-> | 0x03 | 0x02 |
    ---------------

Then you interpret it as an unsigned short:

printf("%d\n%d\n%d\n", e.y[0], e.y[1], e.x);

You’re on a little-endian system (as @Oli noted), meaning the least-significant byte is stored first in memory. Which means that when your code looks at an unsigned short, it thinks 0x03 is the least-significant byte.

So your 2-byte unsigned short is interpreted as 0x0203. And 0x0203 hex is 515 decimal.

---

That comment was interesting enough to put in the answer for clarity, I think.

Let’s say we do this:

union a {
    int x;
    char y[2];
};

int main(int argc, char * argv[])
{
    union a e = {512};
}

What’s inside? Break it down:

int is 4 bytes, char [2] is 2 bytes, so the union is 4 bytes long to store the largest datatype. 512 is 0x00000200 in hex. So store that integer little-endian and you have:

    -----------------------------
e-> | 0x00 | 0x02 | 0x00 | 0x00 |
    -----------------------------

So e.x is 512. e.y[0] is 0 and e.y[1] is 2